Answer:
CH2CH3 - methyl-pentane
CH3CH2-C-CH2CH3 - 2-methyl-4hexyne
CH3- methyl
Answer: 10
Explanation:
The detailed solution is contained in the image attached. The molar mass of hydrated and anhydrous salts are obtained and the number of moles of hydrated and hydrated salts are equated. The masses of hydrated and anhydrous salts are gives in the question and are simply substituted accordingly. This can now be used to obtain the number of molecules of water of crystallization as required in the question.
Answer:
Deshydrohalogenation
Explanation:
You are not providing options to answer, however, this can be answered without options.
Now, in the reaction we can see that we have an atom of Bromine in carbon 2, and in the product appears as NaBr. This means that the Br was substracted by elimination. It's an elimination because the final product do not have a substituent where the bromine was, (Like another nucleophyle such OH or another halide). If you look closely the final product, we can see that one hydrogen in carbon 3, is no longer there. So this electrophyle was also substracted, in this case, by a base (Such NaOH), so in this case, it's ocurring an elimination reaction via E2 (One step, bimolecular). So, as the final product has been substracted the nucleophyle and electrophyle, this treaction is a deshydrohalogenation (an atom of hydrogen and a halide were substracted). The mechanism of this, you can see it in the picture.
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